When does $\sqrt{x+\sqrt{x+1+\sqrt{x+2+...}}}=0$? Consider the function $f$ defined as the limit of the functions
$$f_0(x)=\sqrt{x}$$
$$f_1(x)=\sqrt{x+\sqrt{x+1}}$$
$$f_2(x)=\sqrt{x+\sqrt{x+1+\sqrt{x+2}}}$$
$$...$$
so that $f(x)$ is defined iff $f_n(x)$ is defined for some $n$. The unique root $x_0$ of the function $f$ satisfies $f(x_0)=0$, and it can be alternatively expressed as the limit of the roots of $f_0, f_1, f_2, ...$. See the graphs below:

Can anyone find an expression equal to this limit? I realize that the chances of something nice and closed-form are slim - can we find a series, integral, or even nested radical representation of the real root of $f(x)$?
 A: This problem is equivalent to proving that there exists some real number $r$ such that 
\begin{equation}
\lim_{n \to \infty} f_n(r) = 0 
\end{equation}
where $f_n = g_n^0(x)$ and $g^k_n$ is defined by the recurrence 
\begin{equation}
g_n^k(x) = 
\begin{cases} \sqrt {x+k} & \text{if }n=1 \\ 
\sqrt{x + k +  g^{k+1}_{n-1}(x)}  & \text{o.w.} \\ 
\end{cases}. 
\end{equation}
Therefore, by definition, we have 
\begin{equation}
f_n(x) = 
\begin{cases} \sqrt {x} & \text{if }n=1 \\ 
\sqrt{x +  g^{1}_{n-1}(x)}  & \text{o.w.} \\ 
\end{cases},
\end{equation}
which after proving the following:
\begin{equation}
 g_n^1(x)  = g_n^0(x+1) ,
\end{equation} 
 by a simple induction, gives us that
\begin{equation}
f_n(x) = 
\begin{cases} \sqrt {x} & \text{if }n=1 \\ 
\sqrt{x +  f_{n-1}(x+1)}  & \text{o.w.} \\ 
\end{cases}.
\end{equation}
Let us suppose that $r_n - f_{n-1}(r_n+1) =  \epsilon_n$ and that $\lim_{n \to \infty} \epsilon_n  = 0 $ then we have that 
\begin{equation}
\lim_{n \to \infty} f_{n}(r_n) = 0.
\end{equation} 
One possible route to find this is to consider the growth of the sequence
\begin{equation}
\delta_n(x)  =  f_n(x+1) - f_n(x). 
\end{equation}
Therefore given a good approximation to $r_n$, an educated guess for $r_{n+1} $ should be some
\begin{equation}
r_{n+1} \in [r_{n} - 1, r_{n}],
\end{equation}
this is because
\begin{equation}
f_{n+1}^2(r_{n} - \epsilon) = r_{n} - \epsilon + f_n(r_{n} - \epsilon) +\delta_n(r_{n} - \epsilon)  
\end{equation}
and 
\begin{equation}
\delta_n(r_{n} - \epsilon) =   f_n(r_{n} +1- \epsilon) - f_n(r_{n} - \epsilon) =  f_n(r_{n} +(1-\epsilon)) - f_n(r_{n} - \epsilon)
\end{equation}
gives us that we should be looking for some 
\begin{equation}
-\delta_n(r_{n} - \epsilon)  \approx r_{n} - \epsilon + f_n(r_{n} - \epsilon)
\end{equation}
which is equivalent to saying 
\begin{equation}
-\delta_n(r_{n} - \epsilon)  - f_n(r_{n} - \epsilon) \approx r_{n} - \epsilon 
\end{equation}
or 
\begin{equation}
-f_n(r_{n} + (1- \epsilon))  \approx r_{n} - \epsilon, 
\end{equation}
or 
\begin{equation}
f_n(r_{n} + (1- \epsilon))  \approx \epsilon - r_n .
\end{equation}
which is a "sort-of" fixed point equation. Now we are getting somewhere! 
We now have a method to recursively approximate solutions from previous solutions; i.e.
\begin{equation}
f_n(r_{n} + (1- \epsilon)) =   \epsilon -r_{n}  \implies f_{n+1}(r_{n} - \epsilon) = 0  . 
\end{equation}
Let us compute $r_n$ for small values to verify the solution. 


*

*$f_1(x) = \sqrt x $ is the base case and it has the obvious solution $r_1 = 0$.

*Since $f_1(0) = 0$ and $\sqrt{1- \frac{\sqrt5  -1}{2}} = \frac{\sqrt5  -1}{2} $ we have an exact solution in the form of $r_1 = 0 $, $\epsilon = \frac{\sqrt5  -1}{2} $, and $r_2 = r_1 - \epsilon $; i.e. $f_2 (r_1 - \epsilon) = f_2\left(\frac{ 1-\sqrt5 }{2}\right) = 0$.

*Let $\epsilon = \frac{1}{2} (3 - \sqrt{5})$ then one can easily check that both $f_2(r_2 +1 - \epsilon) = \epsilon - r_2 $ and $f_3(r_2  - \epsilon) =f_3(-1) = 0 . $

*Let $\epsilon = 0.153761 $ then one can easily check that both $f_2(r_3 + 1 - \epsilon) \approx \epsilon - r_3 $ and $f_4(r_3  - \epsilon) = f_3(-1.153761) \approx 0 . $
In conclusion, we have the following recursion relation for the roots

\begin{equation}
r_{n+1} = 
\begin{cases} 0 & \text{if }n=1 \\ 
 r_n - \epsilon  & \text{if } f_n(r_n +1 - \epsilon) =  r_n - \epsilon  \\ 
\end{cases};
\end{equation}

however, it is possible that a recursive analysis of the following two functions 


*

*$f'_n(x)$

*$\delta_n(x)$
could give a better recursive bound $b_n  < b_{n-1}<1 $ on the intervals
\begin{equation}
r_{n+1} \in [r_n-b, r_n]
\end{equation}
A: To address how I performed the computation, this was done in Mathematica using the following code:
F[x_, n_] := Fold[Sqrt[x + #1 + #2] &, 0, Reverse[Range[n + 1] - 1]]
FindRoot[F[x, 500] == 0, {x, -1.2}, WorkingPrecision -> 50]

The first command defines $f_n(x)$; the second chooses $n = 500$, which due to the extremely rapid convergence of $\{f_n\}_{n \ge 1}$, is more than sufficient to converge to the desired precision with a short computation time.  You can check that the result is accurate by choosing $n = 100$ and seeing that the result is unchanged to $50$ digits of precision; indeed, even to $100$ digits of precision.  I would put an upper bound on the error to be less than $10^{-n}$ when using $f_n$ instead of $f$.
A: Not sure if this is helpful, but it may be fruitful to consider a sequence of functions $g_n(x)$ where $g_0(x) = x$ and
$$g_n(x) = [g_{n-1}(x)]^2 - x - n$$
In particular, 
$$\lim_{n \to \infty} g_n(x_0) = 0.$$
I found this by setting
$$f_n(x) = 0$$
and moving all roots to the other side. For example, with $n = 2$,
$$\begin{align*}
0 &= \sqrt{x + \sqrt{x + 1 + \sqrt{x + 2 + \sqrt{x+3}}}}, \\
-x &= \sqrt{x + 1 + \sqrt{x + 2 + \sqrt{x+3}}}, \\
x^2 - x - 1 &= \sqrt{x+2 + \sqrt{x+3}}, \\
(x^2 - x - 1)^2 - x - 2 &= \sqrt{x+3}, \\
((x^2 - x - 1)^2 - x - 2)^2 - x - 3 &= 0.
\end{align*}$$
This makes it clear that if $a$ satisfies $f_n(a) = 0$, then $g_n(a) = 0$. I suppose then that the solution to the question is dependent on if the sequence $g_n$ defined by $g_0 = x$ and 
$$g_n = g_{n-1}^2 - x - n$$
has an explicit form.
A: $\color{brown}{\textbf{The task standing.}}$
Let $-r$ is the root of $f(x).$
Since $f(x)$ is increasing function and 
$$f(-r)=0,\quad f(-1) > f_2(-1) = 0,$$ 
then $r>1.$
At the same time,
\begin{cases}
f(x+1) = f^2(x)-x\\
f(-r)=0\\
f(-r+1) = r\\
f(-r+2) = r^2+r-1\\
f(-r+3)= (r^2+r-1)^2 + r-2 = (r-1)(r^4+3r^2+2r+1)\dots.\tag1
\end{cases}
Besides,
\begin{cases}
f'(x+1) = 2f(x)f'(x)-1\\
f'(-r) = \infty\\
f'(-r+2) = 2rf'(-r+1)-1\\
f'(-r+3) = 2(r^2-r+1)f'(-r+2)-1\dots\tag2
\end{cases}
Conditions $(1),(2)$ allow to define unknown parameters in the various parametric representations of $f(x).$
$\color{brown}{\textbf{Inverse polynomial model.}}$
Let the inverse function is $f^{-1}(x)=g(y),$ then 
\begin{cases}
g(0) = -r\\
g'(0)=0\\
g(r) = -r+1\\
g(r^2+r-1) = -r+2\\
g((r-1)(r^4+3r^2+2r+1)) = -r+3,\tag3
\end{cases}
and the coefficients of the cubic approximating polynomial in the form of
$$g(y)=-r+qy^2+py^3\tag4$$
can be obtained from the algebraic system
\begin{cases}
-r+1 = -r + pr^2 + qr^3\\
-r+2 = -r + p(r^2+r-1)^2 + q(r^2+r-1)^3\\
-r+3 = -r + p(r-1)^2(r^4+3r^2+2r+1)^2 + q(r-1)^3(r^4+3r^2+2r+1)^3,
\end{cases}
with the single positive solution
$$p\approx 0.0622998,\quad q\approx 0.606587,\quad r\approx1.21088$$
(see also Wolfram Alpha results).
Therefore, the cubic model estimation of root of $f(x)$ is $\color{brown}{\mathbf{-1.21088}}.$
$\color{brown}{\textbf{Power model.}}$
Taking in account conditions $(1),$ can be used the explicit power model
$$f(x) = r(x+r)^p,\tag5$$
where
\begin{cases}
r^2+r-1 = r\cdot 2^p\\
(r-1)(r^3+3r^2+2r+1) = r\cdot 3^p,
\end{cases}

The power model estimation of root of $f(x)$ is $\color{brown}{\mathbf{-1.21168}}.$
Plot of the inverse functions for the considered models see below.

Using of detalized models can improve approximations accuracy.
A: Since there is apparently no representation of the root of $f(x)$ as a series, integral, or nested radical, this answer will explain how to find the root by numerically solving $f(x)=0.$  Of course, if you have Mathematica, you could use the Mathematica code in heropup's answer.  But the Mathematica code doesn't explain exactly how the root is computed.  Since the derivative of $f(x)$ is infinite at the root $x=r$ and $f$ is undefined for $x\lt r,$ neither Newton's method nor the secant method work.  But the bisection method will work if you make a slight change in the definition of $f(x)$: if any argument of a square root during the computation of $f(c)$ is negative, define the value of $f(c)$ as negative one.

The function $f(x)$ can be generalized by replacing the sequence $1, 2, 3, \dots$ with the more general arithmetic sequence $C, 2C, 3C, \dots$ (where $C$ is some non-negative real number).  The following C code computes the root of $f_N(x,C)$ in quad precision arithmetic.  It computes the roots of $f_{50}(x,1)$, $f_{50}(x,2)$, and $f_{50}(x,3)$ as $-1.21103728351247151496696981615$, $-1.62419058855232395793631994913$, and $-1.94153569702169327219376232801$.

// Compute f(x,N,C).  
//
// f(x,0,C) = sqrt(x),
// f(x,1,C) = sqrt(x + sqrt(x+C)),
// f(x,2,C) = sqrt(x + sqrt(x+C + sqrt(x+2C))), ...
//
// Return -1 if any argument of sqrt is negative.
//
__float128 f(__float128 x, int N, __float128 C)
{
  __float128 fval=0;
  for (int i=0; i<=N; i++) {
    fval = x + C*(N-i) + fval;
    if (fval < 0) return -1;
    fval = sqrtq(fval);
  }

  return fval;
}

int main(int argc, char **argv)
{
  char f128_s[256];

  if (argc < 3) {
    printf("Usage: (N) (C)\n");
    return 1;
  }
  int N = atoi(argv[1]);
  __float128 C = strtoflt128(argv[2], NULL);
  if (N < 2) {
    printf("N must be at least 2\n");
    return 1;
  }

  // Solve f(x,N,C) = 0 for x using bisection method.  Exit bisection
  // method loop when maximum number of iterations is exceeded or
  // length of interval containing root is smaller than specified
  // tolerance.
  __float128 x0=-100, x1=0;
  int maxit=150, numit=1; 
  __float128 min_interval_len=2e-32, root;
  while ( 1 ) {
    if (numit >= maxit) break;

    root = (x0 + x1)*0.5;
    if (x1-x0 < min_interval_len) break;

    __float128 fval = f(root,N,C);

    if (fval > 0) x1 = root;
    else x0 = root;

    numit++;
  }

  quadmath_snprintf(f128_s, 256, "%.30Qg", root);
  printf("root=%s, numit=%d\n", f128_s, numit);
}

